A curve has a parametric equation x=〖at〗^2 and y=2at. If the area is bounded by the curve between t=1 and t=2
If you mean the area under the curve, that is, as usual,
∫ y dx = ∫[1,2] 2at * (2at * 2a) dt = ∫[1,2] 8a^3 t^2 dt = 56/3 a^3
oops. did you see my typo? Better fix it.
To find the area bounded by the curve between t=1 and t=2, we need to calculate the definite integral of y with respect to x over this interval.
First, let's express y as a function of x using the given parametric equations:
From the equation x = at^2, we can solve for t as follows:
t^2 = x/a
t = √(x/a)
Substituting this value of t into the equation y = 2at, we get:
y = 2a√(x/a)
y = 2√(ax)
Now, the area under the curve can be found by integrating y with respect to x from x = a to x = 4a for the given range of t.
∫[1,2] (2√(ax)) dx
= 2 ∫[1,2] √(ax) dx
To solve this integral, we can use a substitution. Let's substitute u = √(ax), then du/dx = √(a/x), and dx = (x/a) du.
The integral becomes:
2 ∫[1,2] u (x/a) du
Now, we need to express the limits of integration in terms of u. For x = 1, u = √(a) and for x = 2, u = √(2a).
Using the substitution and converting the limits of integration, we have:
2 ∫[√(a),√(2a)] u (x/a) du
Simplifying further:
(2/a) ∫[√(a),√(2a)] x du
Integrating x with respect to u gives:
(2/a) ∫[√(a),√(2a)] (a u^2) du
= (2/a) [a(u^3)/3] |[√(a),√(2a)]
= (2/a) [(a(√(2a))^3)/3 - (a(√(a))^3)/3]
= (2/3a) (√(2a)^3 - √(a)^3)
Simplifying further, we get:
= (2/3a) [2√(a^3) - √(a^3)]
= (2/3a) (√(a^3))
= 2/3√(a)
Therefore, the area bounded by the curve between t=1 and t=2 is 2/3√(a).
To find the area bounded by the curve between t=1 and t=2, we can use the concept of definite integrals.
First, we need to express the area as an integral. The formula for the area between curves is:
Area = ∫[a, b] (f(x) - g(x)) dx
In this case, the curve is defined parametrically as x=at^2 and y=2at. We can rewrite x in terms of y using the parametric equations:
x = (y/2a)^2
Now, we need to find the bounds for integration. Since t=1 and t=2 are given, we substitute these values into the parametric equations to find the corresponding x and y values:
For t=1:
x = a(1^2) = a
y = 2a(1) = 2a
For t=2:
x = a(2^2) = 4a
y = 2a(2) = 4a
Now, we have two points on the curve: (a, 2a) and (4a, 4a).
To find the area, we need to find the difference between the upper curve, y=4a, and the lower curve, y=2a, and integrate it with respect to x. We can rewrite the curves in terms of x:
Upper curve: y = 4a = 2x
Lower curve: y = 2a = x/2
Substituting these curves into the area formula, we have:
Area = ∫[a, 4a] (2x - x/2) dx
Now, we can simplify the integrand:
Area = ∫[a, 4a] (3x/2) dx
Integrating, we get:
Area = [(3/2) * (x^2/2)]|[a, 4a]
Simplifying further, we have:
Area = (3/4) * (16a^2 - a^2)
Area = (3/4) * 15a^2
Finally, we can simplify the expression to:
Area = (45/4) * a^2
So, the area bounded by the curve between t=1 and t=2 is (45/4) * a^2.